# Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 3

## Theorem

Let $\sequence {x_n}$ be the real sequence defined as $x_n = \paren {n + 1}^{1/n}$, using exponentiation.

Then $\sequence {x_n}$ converges with a limit of $1$.

## Proof

We have the definition of the power to a real number:

$\paren {n + 1}^{1/n} = \map \exp {\dfrac 1 n \, \map \ln {n + 1} }$

For $n >= 1$ then $n + 1 \le 2 n$.

Hence:

 $\displaystyle \frac 1 n \, \map \ln {n + 1}$ $\le$ $\displaystyle \frac 1 n \, \map \ln {2 n}$ Logarithm is Strictly Increasing $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \frac 1 n \paren {\ln 2 + \ln n}$ Logarithm on Positive Real Numbers is Group Isomorphism $\displaystyle$  $\displaystyle$ $\displaystyle$ $=$ $\displaystyle \frac {\ln 2} n + \frac 1 n \ln n$
$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n \ln n = 0$
$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n = 0$
$\displaystyle \lim_{n \mathop \to \infty} \paren {\frac {\ln 2} n + \frac 1 n \ln n} = \ln 2 \cdot 0 + 0 = 0$
$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} = 0$

Hence:

$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} = \exp 0 = 1$

and the result follows.

$\blacksquare$